Evaluate $\int_0^1 \int_0^{1-y} \cos \left(\frac{x-y}{x+y}\right)dxdy$ 
Evaluate the following double integral
$$\int_0^1 \int_0^{1-y} \cos \left(\frac{x-y}{x+y}\right)dxdy$$

I tried transforming to
\begin{align}
x+y &=u\\
x-y &=v
\end{align} but I think it is getting complicated.  Thanks in advance.
 A: You are on the right track. By letting $u=x+y$ , $v=x-y$, we have
$$\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\left|\frac{\partial (u,v)}{\partial (x,y)}\right|^{-1}=\left|\det \begin{pmatrix}
1 &  1 \\   1 &  -1
\end{pmatrix}\right|^{-1}=|-2|^{-1}=\frac{1}{2}.$$
Moreover the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$ in the $xy$-plane is transformed into the triangle with vertices $(0,0)$, $(1,1)$ and $(1,-1)$ in the $uv$-plane.
Therefore
$$\int_{y=0}^1\left( \int_{x=0}^{1-y} \cos\left(\frac{x-y}{x+y}\right)dx\right)dy=\frac{1}{2}\int_{u=0}^1\left(\int_{v=-u}^u\cos(v/u)dv\right) du.$$
Can you take it from here?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\int_{0}^{1}\int_{0}^{1 - y}\cos\pars{x - y \over x + y}\,\dd x\,\dd y
\,\,\,\stackrel{x\ \mapsto\ \pars{1 - y} - x}{=}\,\,\,
\int_{0}^{1}\int_{0}^{1 - y}\cos\pars{1 - x - 2y \over 1 - y - x + y}
\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1 - y}\cos\pars{1 - {2y \over 1 - x}}\,\dd x\,\dd y =
\int_{0}^{1}\int_{0}^{1 - x}\cos\pars{1 - {2y \over 1 - x}}\,\dd y\,\dd x
\\[5mm] \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, &
\int_{0}^{1}\int_{0}^{x}\cos\pars{1 - {2y \over x}}\,\dd y\,\dd x =
\int_{0}^{1}\bracks{-\,{1 \over 2}\,x\,\sin\pars{1 - {2y \over x}}}
_{\ y\ =\ 0}^{\ y\ =\ x}\,\,\,\,\dd x
\\[5mm] = &\
-\,{1 \over 2}\int_{0}^{1}x\bracks{\sin\pars{-1} - \sin\pars{1}}\,\dd x =
\sin\pars{1}\int_{0}^{1}x\,\dd x =
\bbx{{1 \over 2}\,\sin\pars{1}} \approx 0.4207
\end{align}
